Question

In a triangle PQR,PQ > PR and S is a point equidistant from the three sides of the triangle. Prove that SQ>SR

Answer 1

In ∆PQR, we have, PQ > PR given⇒ ∠PRQ > ∠PQR angle opposite to longer side of a ∆ is greater⇒12∠PRQ > 12∠PQR ........1Since, SR bisects ∠R, then∠SRQ = 12∠PRQ ........2Since SQ bisects ∠P, then∠SQR = 12∠PQR .......3Now, from 1, we have 12∠PRQ > 12∠PQR⇒∠SRQ > ∠SQR using 2 and 3Now, in ∆SQR, we have ∠SRQ > ∠SQR proved above⇒ SQ > SR i.e. side opposite to greater angle of a ∆ is longer